Question
Asked 24th Feb, 2015

Does the braking distance of a car depend on weight of the car?

d = v2/2ug, 
d = minimum stopping distance,
v = velocity when brakes were applied
u = coeff. of friction between wheels and road
g = gravitational constant,
The above equation shows that braking distance is independent of mass of vehicle. I know that depending on the value of slip, the value of 'u' will change, but lets assume that 'u' is constant. So does it mean that two different vehicles, one a normal car and other a sports car would have a same braking distance for a given same 'v'. I am confused because this does not seem to be the case. What am I missing?

Most recent answer

Reza Dariani
Hochschule Merseburg

All Answers (8)

Sascha Eisenträger
Otto-von-Guericke-Universität Magdeburg
Hi,
you could try to use D'Alemberts principle or the Energy balance to solve this dynamical problem.
If you use the energy balance then you compare to states. In the first state the car has only kinetic energy E1 = 1/2*m*v^2. Since the car breaks until standstill it has no kinetic energy in the second state. Due to the friction, however, the system is not conservative. The kinetic energy of the first state is consumed by the work of the friction force E2 = umgd. The energy balance now gives you
E1 = E2
And you get d = v2/2ug.
Best regards,
Badril Hisham Abu bakar
Malaysian Agricultural Research and Development Institute (MARDI)
The weight does play a role. u = F_friction/F_normal. F_normal is the force that presses the road and tyres together, i.e. the weight of the car, F_normal = mg.
If we are comparing two cars, in your case a sedan car and a sports car, assuming that the mass is different for both cars (m_sports > m_sedan), say:
g = 10N/kg
m_sports = 2kg, F_sports = 2kg * g = 20N
m_sedan = 1kg, F_sedan = 1kg * g = 10N
say u = 1 for both cars,
then the only way u can be constant is by F_friction of both cars to change accordingly, 
u_sedan = F_friction/F_sedan = 10N/10N = 1
u_sports = F_friction/F_sports = 20N/20N = 1
i.e. if the sports car has bigger mass, then the tyres need to be bigger as well. So if we say that u is constant, we are actually saying we are compensating for the mass of the cars with frictional force.
In reality though, for the same v, two cars (sports and sedan)  do not have the same u, that is why you will see one go off a cliff.
Sascha Eisenträger
Otto-von-Guericke-Universität Magdeburg
For the sake of Argument we assume that u is equal in both cases and since the frictional force only depends on the normal force and u the mass does not Play a role theoretically.
Even in reality that should be the case if u is constant and acceleration is the same for both cars.
Best,
1 Recommendation
Ramlan Kasiran
Universiti Teknologi MARA
Theoretically it is but practically it is not. This is because of the varying operating variables such as normal loads (unladen and laden state) and load transfer effects during braking. As such, we can refer to EEC Directive 75/524 as guide for brake efficiency  and adhesion utilisation analyses in order to incorporate suitable brake bias or proportioning valves to ensure good efficiency or adhesion utilisation trough out the possible vehicle operating variables.        
Alessandro Vigliani
Politecnico di Torino
You are right: theoretically, braking distance does not depend on mass. But in practice, yes. The explanation is that tire effective friction coefficient cannot be assumed constant: it slightly decreases with increasing normal force. So braking distances are longer when more mass is added. As a limit example, consider that on trucks, the tire coefficient usually is smaller than 0.65 even on dry roads>
please read the answer in attach file
Reza Dariani
Hochschule Merseburg

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How can I estimate the difference between two rotation matrices?
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  • Dennis RobertDennis Robert
The main objective here is to computer pose of a real camera in the subsequent frames of a video (after I compute it in the first frame using a PnP algorithm) without the need of solving PnP again and again.
I have two rotation matrices. One is that of the rotation matrix of a real webcam which I got by solving the PnP problem. I have a world coordinate frame and I know the locations of each and every point in world coordinates in the world space.
As far as I understand, a rotation matrix transforms points in world coordinates to camera frame coordinates (not considering translation here). This means that, R1 gives you the orientation of world coordinate frame with respect to camera coordinate frame.
My second rotation matrix is that of a sensor which is also in world coordinates. ie, this rotation matrix gives you the orientation of world coordinate frame with respect to sensor coordinate frame.
I want to find the orientation of real webcam coordinate frame with respect to sensor coordinate frame.
Lets name the first rotation matrix Rw1 and the second rotation matrix Rw2, with the subscripts w1 denoting world with respect to real webcam and w2 denoting world with respect to sensor (1 and 2 can be considered denoting real webcam and sensor respectively).
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Let's consider the second frame for now.
I know R12 (which I am assuming is constant and I computed it in first frame) and Rw2 (sensor rotation matrix for the second frame). I have to find Rw1 for the second frame.
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And then I repeat the same process for all the subsequent frames using the above formula. The only thing which is changing is Rw2 and I have it for all the frames. R12 matrix will not be changing according to my assumption.
Is my method and assumption valid ?
PS: (R)' means the transpose of R.

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